Integrand size = 24, antiderivative size = 104 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^3(d+b x) \, dx=\frac {2 e^{a-d}}{b \left (1-e^{2 d+2 b x}\right )}-\frac {2 e^{a-d}}{b \left (1+e^{2 d+2 b x}\right )^2}+\frac {4 e^{a-d}}{b \left (1+e^{2 d+2 b x}\right )}-\frac {2 e^{a-d} \text {arctanh}\left (e^{2 d+2 b x}\right )}{b} \] Output:
2*exp(a-d)/b/(1-exp(2*b*x+2*d))-2*exp(a-d)/b/(1+exp(2*b*x+2*d))^2+4*exp(a- d)/b/(1+exp(2*b*x+2*d))-2*exp(a-d)*arctanh(exp(2*b*x+2*d))/b
Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^3(d+b x) \, dx=\frac {2 e^{a-d} \left (\frac {1}{1-e^{2 (d+b x)}}-\frac {1}{\left (1+e^{2 (d+b x)}\right )^2}+\frac {2}{1+e^{2 (d+b x)}}-\text {arctanh}\left (e^{2 (d+b x)}\right )\right )}{b} \] Input:
Integrate[E^(a + b*x)*Csch[d + b*x]^2*Sech[d + b*x]^3,x]
Output:
(2*E^(a - d)*((1 - E^(2*(d + b*x)))^(-1) - (1 + E^(2*(d + b*x)))^(-2) + 2/ (1 + E^(2*(d + b*x))) - ArcTanh[E^(2*(d + b*x))]))/b
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.62, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2720, 27, 354, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{a+b x} \text {csch}^2(b x+d) \text {sech}^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {32 e^{a+5 b x}}{\left (1-e^{2 b x}\right )^2 \left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {32 e^a \int \frac {e^{5 b x}}{\left (1-e^{2 b x}\right )^2 \left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {16 e^a \int \frac {e^{2 b x}}{\left (1-e^{2 b x}\right )^2 \left (1+e^{2 b x}\right )^3}de^{2 b x}}{b}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {16 e^a \int \left (-\frac {1}{4 \left (1+e^{2 b x}\right )^2}+\frac {1}{4 \left (1+e^{2 b x}\right )^3}+\frac {1}{8 \left (-1+e^{2 b x}\right )}+\frac {1}{8 \left (-1+e^{2 b x}\right )^2}\right )de^{2 b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {16 e^a \left (-\frac {1}{8} \text {arctanh}\left (e^{2 b x}\right )+\frac {1}{8 \left (1-e^{2 b x}\right )}+\frac {1}{4 \left (e^{2 b x}+1\right )}-\frac {1}{8 \left (e^{2 b x}+1\right )^2}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Csch[d + b*x]^2*Sech[d + b*x]^3,x]
Output:
(16*E^a*(1/(8*(1 - E^(2*b*x))) - 1/(8*(1 + E^(2*b*x))^2) + 1/(4*(1 + E^(2* b*x))) - ArcTanh[E^(2*b*x)]/8))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 0.26 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.45
\[\frac {2 \left (-{\mathrm e}^{4 b x +4 a +4 d}+3 \,{\mathrm e}^{2 b x +4 a +2 d}+2 \,{\mathrm e}^{4 a}\right ) {\mathrm e}^{3 a -d}}{\left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) \left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}-\frac {\ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{a -d}}{b}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}-{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{a -d}}{b}\]
Input:
int(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^3,x)
Output:
2/(-exp(2*b*x+2*a+2*d)+exp(2*a))/(exp(2*b*x+2*a+2*d)+exp(2*a))^2/b*(-exp(4 *b*x+4*a+4*d)+3*exp(2*b*x+4*a+2*d)+2*exp(4*a))*exp(3*a-d)-ln(exp(2*b*x+2*a )+exp(2*a-2*d))/b*exp(a-d)+ln(exp(2*b*x+2*a)-exp(2*a-2*d))/b*exp(a-d)
Leaf count of result is larger than twice the leaf count of optimal. 1384 vs. \(2 (94) = 188\).
Time = 0.11 (sec) , antiderivative size = 1384, normalized size of antiderivative = 13.31 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^3(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^3,x, algorithm="fricas")
Output:
(2*cosh(b*x + d)^4*cosh(-a + d) + 2*(cosh(-a + d) - sinh(-a + d))*sinh(b*x + d)^4 + 8*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh (b*x + d)^3 - 6*cosh(b*x + d)^2*cosh(-a + d) + 6*(2*cosh(b*x + d)^2*cosh(- a + d) - (2*cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh(-a + d))*sinh(b*x + d )^2 - (cosh(b*x + d)^6*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b *x + d)^6 + 6*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*si nh(b*x + d)^5 + cosh(b*x + d)^4*cosh(-a + d) + (15*cosh(b*x + d)^2*cosh(-a + d) - (15*cosh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d))*sinh(b*x + d )^4 + 4*(5*cosh(b*x + d)^3*cosh(-a + d) + cosh(b*x + d)*cosh(-a + d) - (5* cosh(b*x + d)^3 + cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d)^3 - cosh(b*x + d)^2*cosh(-a + d) + (15*cosh(b*x + d)^4*cosh(-a + d) + 6*cosh(b*x + d)^2 *cosh(-a + d) - (15*cosh(b*x + d)^4 + 6*cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh(-a + d))*sinh(b*x + d)^2 + 2*(3*cosh(b*x + d)^5*cosh(-a + d) + 2*co sh(b*x + d)^3*cosh(-a + d) - cosh(b*x + d)*cosh(-a + d) - (3*cosh(b*x + d) ^5 + 2*cosh(b*x + d)^3 - cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d) - (cos h(b*x + d)^6 + cosh(b*x + d)^4 - cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh( -a + d))*log(2*cosh(b*x + d)/(cosh(b*x + d) - sinh(b*x + d))) + (cosh(b*x + d)^6*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b*x + d)^6 + 6*(c osh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^5 + cosh(b*x + d)^4*cosh(-a + d) + (15*cosh(b*x + d)^2*cosh(-a + d) - (15*c...
\[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^3(d+b x) \, dx=e^{a} \int e^{b x} \operatorname {csch}^{2}{\left (b x + d \right )} \operatorname {sech}^{3}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*csch(b*x+d)**2*sech(b*x+d)**3,x)
Output:
exp(a)*Integral(exp(b*x)*csch(b*x + d)**2*sech(b*x + d)**3, x)
Time = 0.05 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.55 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^3(d+b x) \, dx=-\frac {e^{\left (a - d\right )} \log \left (e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} + e^{\left (2 \, a\right )}\right )}{b} + \frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right )}{b} + \frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} - e^{a}\right )}{b} + \frac {2 \, {\left (e^{\left (4 \, b x + 7 \, a + 4 \, d\right )} - 3 \, e^{\left (2 \, b x + 7 \, a + 2 \, d\right )} - 2 \, e^{\left (7 \, a\right )}\right )}}{b {\left (e^{\left (6 \, b x + 6 \, a + 7 \, d\right )} + e^{\left (4 \, b x + 6 \, a + 5 \, d\right )} - e^{\left (2 \, b x + 6 \, a + 3 \, d\right )} - e^{\left (6 \, a + d\right )}\right )}} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^3,x, algorithm="maxima")
Output:
-e^(a - d)*log(e^(2*b*x + 2*a + 2*d) + e^(2*a))/b + e^(a - d)*log(e^(b*x + a + d) + e^a)/b + e^(a - d)*log(e^(b*x + a + d) - e^a)/b + 2*(e^(4*b*x + 7*a + 4*d) - 3*e^(2*b*x + 7*a + 2*d) - 2*e^(7*a))/(b*(e^(6*b*x + 6*a + 7*d ) + e^(4*b*x + 6*a + 5*d) - e^(2*b*x + 6*a + 3*d) - e^(6*a + d)))
Time = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.26 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^3(d+b x) \, dx=-\frac {1}{2} \, {\left (\frac {2 \, e^{\left (-6 \, d\right )} \log \left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}{b} - \frac {2 \, e^{\left (-6 \, d\right )} \log \left ({\left | e^{\left (2 \, b x + 2 \, d\right )} - 1 \right |}\right )}{b} + \frac {2 \, {\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )} e^{\left (-6 \, d\right )}}{b {\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}} - \frac {{\left (3 \, e^{\left (4 \, b x + 4 \, d\right )} + 14 \, e^{\left (2 \, b x + 2 \, d\right )} + 7\right )} e^{\left (-6 \, d\right )}}{b {\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{2}}\right )} e^{\left (a + 5 \, d\right )} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^3,x, algorithm="giac")
Output:
-1/2*(2*e^(-6*d)*log(e^(2*b*x + 2*d) + 1)/b - 2*e^(-6*d)*log(abs(e^(2*b*x + 2*d) - 1))/b + 2*(e^(2*b*x + 2*d) + 1)*e^(-6*d)/(b*(e^(2*b*x + 2*d) - 1) ) - (3*e^(4*b*x + 4*d) + 14*e^(2*b*x + 2*d) + 7)*e^(-6*d)/(b*(e^(2*b*x + 2 *d) + 1)^2))*e^(a + 5*d)
Timed out. \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^3(d+b x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x}}{{\mathrm {cosh}\left (d+b\,x\right )}^3\,{\mathrm {sinh}\left (d+b\,x\right )}^2} \,d x \] Input:
int(exp(a + b*x)/(cosh(d + b*x)^3*sinh(d + b*x)^2),x)
Output:
int(exp(a + b*x)/(cosh(d + b*x)^3*sinh(d + b*x)^2), x)
Time = 0.25 (sec) , antiderivative size = 313, normalized size of antiderivative = 3.01 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^3(d+b x) \, dx=\frac {e^{a} \left (e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}-1\right )+e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}+1\right )-e^{6 b x +6 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-2 e^{6 b x +6 d}+e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )+e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )-e^{4 b x +4 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )-e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )+e^{2 b x +2 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-4 e^{2 b x +2 d}-\mathrm {log}\left (e^{b x +d}-1\right )-\mathrm {log}\left (e^{b x +d}+1\right )+\mathrm {log}\left (e^{2 b x +2 d}+1\right )-2\right )}{e^{d} b \left (e^{6 b x +6 d}+e^{4 b x +4 d}-e^{2 b x +2 d}-1\right )} \] Input:
int(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^3,x)
Output:
(e**a*(e**(6*b*x + 6*d)*log(e**(b*x + d) - 1) + e**(6*b*x + 6*d)*log(e**(b *x + d) + 1) - e**(6*b*x + 6*d)*log(e**(2*b*x + 2*d) + 1) - 2*e**(6*b*x + 6*d) + e**(4*b*x + 4*d)*log(e**(b*x + d) - 1) + e**(4*b*x + 4*d)*log(e**(b *x + d) + 1) - e**(4*b*x + 4*d)*log(e**(2*b*x + 2*d) + 1) - e**(2*b*x + 2* d)*log(e**(b*x + d) - 1) - e**(2*b*x + 2*d)*log(e**(b*x + d) + 1) + e**(2* b*x + 2*d)*log(e**(2*b*x + 2*d) + 1) - 4*e**(2*b*x + 2*d) - log(e**(b*x + d) - 1) - log(e**(b*x + d) + 1) + log(e**(2*b*x + 2*d) + 1) - 2))/(e**d*b* (e**(6*b*x + 6*d) + e**(4*b*x + 4*d) - e**(2*b*x + 2*d) - 1))